 Solution

For the free basis $X=\{x_0,x_1,x_2,\dots \}$, defineThe universal property of a free group extends this assignment to an endomorphism of $F(X)$. It is surjective because every $x_n$ is the image of $x_{n+1}$, but it is not injective because $x_0\ne 1$ lies in its kernel. Thus $F(\mathbb{N})$ is not Hopfian.